# Why not solve the quadratic this way?

• Apr 18th 2011, 07:26 PM
bournouli
Why not solve the quadratic this way?
I have (289^2)+(-1666)+(2401)=(x^2-5)

then I, (289^2)+((-1666)+(2401))=(x^2-5)

then I have, (289^2)+(735)=(x^2-5) then subtract (x^2) and add 5 to both sides.

So now I have (288x^2)+740=0

So then, subtract 740 from both parts to get (288x^2)=-740

Finally do this -740/288=x^2, so x=(-2.5)^(1/2) which is imaginary. But the equation has only real solutions. But it looks like the answer was derived using all correct principles...?:confused:
• Apr 18th 2011, 07:34 PM
TheChaz
Quote:

Originally Posted by bournouli
I have (289^2)+(-1666)+(2401)=(x^2-5)

then I, (289^2)+((-1666)+(2401))=(x^2-5)

then I have, (289^2)+(735)=(x^2-5) then subtract (x^2) and add 5 to both sides.

So now I have (288x^2)+740=0

So then, subtract 740 from both parts to get (288x^2)=-740

Finally do this -740/288=x^2, so x=(-2.5)^(1/2) which is imaginary. But the equation has only real solutions. But it looks like the answer was derived using all correct principles...?:confused:

Why are there so many PARENTHESIS?!?!
Is the original equation 289^2 - 1666 + 2401 = x^2 - 5 ???
• Apr 18th 2011, 08:32 PM
Prove It
Maybe because 289^2 doesn't have an x^2 term, so they are not like terms and can not be added/subtracted...
• Apr 18th 2011, 10:10 PM
bournouli
Quote:

Originally Posted by Prove It
Maybe because 289^2 doesn't have an x^2 term, so they are not like terms and can not be added/subtracted...

Sorry it was (289x^2).
• Apr 18th 2011, 10:36 PM
earboth
Quote:

Originally Posted by bournouli
Sorry it was (289x^2).

If and only if the original equation was:

289x^2 - 1666x + 2401 = x^2 - 5

then you'll get the solutions x = 3 or x = 401/144

... but my crystal sphere is a little bit dusty